Question

A body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1cm$. If the angular velocity is doubled, the elongation in the string is $5cm$. The original length of the spring is

• A. 16 cm
• B. 15 cm
• C. 14 cm
• D. 13 cm

Answer 1

Answer: (B)

Let the length of the spring is $l$.
When the system is whirled round in a horizontal circle the centripetal force is given by
$F=\frac{m{v}^2}{r}=\frac{m(r\omega {)}^2}{r}=mr{\omega }^2$
Then, $r=l+$ elongation
Given : elongation $=1cm$ (in the first case)
For angular velocity $\omega$ the force required is
$F_1=m(l+1){\omega }^2=kx=k\times 1=k$
or $k=m(l+1){\omega }^2$...(i)
For second case,${\omega }^{\prime }=2\omega$ , elongation $=5cm=x$ Radius, $r=l+s$
So, $F_2=m(1+5)(2\omega {)}^2=kx=k\times 5=5k$
or $5k=4m(l+5){\omega }^2$ ..(ii)
Now, dividing Eq. (i) by Eq. (ii), we get
$\frac{k}{5k}=\frac{m(l+1){\omega }^2}{4m(l+5){\omega }^2}$
$\Rightarrow 5(l+1)=4(l+5)$
$\Rightarrow 5l+5=4l+20$
$l=20-5=15cm$